 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s25/2/b/solution

λ^{'} = 0

. A decaying normal mode

e^{- k t}

obeys

C_{s} C_{d} k^{2} - [(λ + γ) C_{d} + γ C_{s}] k + λγ = 0.

For

C_{d} ≫ C_{s}

, its sum and product of roots give

k_{1} ≃ \frac{λ + γ}{C _{s}}, k_{2} ≃ \frac{λγ}{C _{d} ( λ + γ )} .

Initially

T_{s} = T_{d} = T_{0}

, so

a_{1} + a_{2} = 1

and

a_{1} k_{1} + a_{2} k_{2} = λ / C_{s}

. To leading order,

a_{1} ≃ \frac{λ}{λ + γ}, a_{2} ≃ \frac{γ}{λ + γ} .

On the slow normal mode, the first equation gives

T_{d} / T_{s} = (λ + γ - C_{s} k_{2}) / γ ≃ (λ + γ) / γ

, verifying proportionality after the fast transient has decayed. The small parameter must also make

k_{2} / k_{1}

small; the supplied positive coefficients and

C_{d} ≫ C_{s}

satisfy this.

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